Angle E 0398F5

1. **State the problem:** We are given a triangle $DEF$ with $m\angle F = 61^\circ$, side $DE = 29$, and side $DF = 23$. We need to find the measure of angle $E$ to the nearest tenth. 2. **Recall the Law of Sines:** The Law of Sines states: $$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$ where $A$, $B$, and $C$ are angles opposite sides $a$, $b$, and $c$ respectively. 3. **Assign sides and angles:** - $m\angle F = 61^\circ$ - Side opposite $\angle E$ is $DF = 23$ - Side opposite $\angle F$ is $DE = 29$ 4. **Use Law of Sines to find $m\angle E$:** $$\frac{\sin E}{DF} = \frac{\sin F}{DE}$$ Substitute known values: $$\frac{\sin E}{23} = \frac{\sin 61^\circ}{29}$$ 5. **Solve for $\sin E$:** $$\sin E = 23 \times \frac{\sin 61^\circ}{29}$$ Calculate $\sin 61^\circ$: $$\sin 61^\circ \approx 0.8746$$ So, $$\sin E = 23 \times \frac{0.8746}{29} = 23 \times 0.03016 = 0.6937$$ 6. **Find $m\angle E$:** $$m\angle E = \sin^{-1}(0.6937) \approx 43.9^\circ$$ 7. **Answer:** The measure of angle $E$ to the nearest tenth is $43.9^\circ$.